priti
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Posts posted by priti
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As you said
"btw its a joomla component so isn't there an already premade function for this?"
then you need to look in to the code of joomla CMS or you can post the same query in joomla forum you will get prompt help !!! :-)
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$recordSet=mysql_query(select * from tablename);
while($row=mysql_fetch_assoc($recordSet))
{
echo $author =$row['author'];
}
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In javascript function
document.form1.submit(); - > It will submit the form1 where which is your submit1
document.form2.submit() : for submit2
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then you can set cookies with help of javascript on page event Onload
one more thing you can try:
place setcookie() function on start of page. "This requires that you place calls to this function prior to any output, including <html> and <head> tags as well as any whitespace. "
Hope it helps
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select * from tablename where ORDER BY word ASC
let us know if this solve the problem in better way.
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Can you try array_push() function in php ?
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It is called FAVICON
Following link will give you plenty of knowledge about it.
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try this
echo "<b>$username</b><br> $state<br> <a href='$url'>View Profile</a><br><br>";
header("Content-type:image/jpeg");
<img src ="$imagelocation" width='100' height='100' border='0'/>
Do let me know error if any error pops up
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before displaying image do
header("Content-type:image/jpeg"); please take care of image format.
and see if it displays
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<img src ="$imagelocation" width='100' height='100' border='0'>
you are missing with " " around $imagelocation
what is the error!!!
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make sure
> query is correctly generated
> query is returning correct output
echo "
<b>$username</b><br>
$state<br>
<a href='$url'>View Profile</a><br><br>
// <------------echo $imagelocation here
";
in echo you are echoing image location ?????
echo "$username-$state-$imagelocation"
what does above line prints. Please make sure that your query returning correct data.
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what is stored in database in imagelocation? can you post example?
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sorry
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what is the error you are receiving and please point where in you are trying to echo $imagelocation in code?
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HTTP Error 403 is an error message which means that you are not authorized to view the web page which you are attempting to load.
check if you are trying to access password protect folder or some functionality which are of type "only authorized user can access it"
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Wish you must have shown us line 80 code it make one to debug your code around that line. !!
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$query="UPDATE nav SET linkname = '" . $_GET['linkname'] . "' , url = '" . $_GET['url'] . "' WHERE id = " . $_GET['id'] . "
echo $query; // run this query in mysql directly.
try this
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webservd root resembles owner-group as you are root you should be able to set change the file permission.
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what is the error you are facing?
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you are receiving that error because allow_url_fopen is OFF in PHP configuration (php.ini) file.
To understand php.ini follow :http://www.slideshare.net/pritisolanki/understanding-phpini-presentation
you can't do ini_set() to set this this is System level configuration value.
I see you are trying to include <?php include("http://justinledelson.com/includes/sidebar.php");?>
if http://justinledelson.com is same site you are working then I think you should do as follow
<?php include("/includes/sidebar.php");?>
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It's not a good practice to name variable like this
if its a delete button then you should name it as - btnDelete
if some Variable is there then it should be - varName
do u get some error if you use document.photocomment.delete.style.visibility ... something like this ?
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'UPDATE nav SET linkname = "$_GET['linkname']" , link= "$_GET['linkname']" WHERE id = $_GET['id'] LIMIT 1;'
will this will be helpful?
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what $abbreviation is showing ??? I amazed that you are not fetching data from $data array and you are not using this array to fetch the data. Can you print what is the output of print_r($data)???
Refer :http://in2.php.net/manual/en/function.fgetcsv.php for more details
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I didn't understand but is following is your requirement
link 1 :yahoo http://www.yahoo.com
link 2:blog http://www.dummy.com
submit button
So when you change yahoo to gmail and link to www.gmail.com in database Link 1 sud get update to gmail-linkname and http://www.gmail.com as link (right)
If above is the scenario then i recommend you to use 'update button' in front of all link set i.e
link1 yahoo link update link
link2 gmail link update link
so when you click update button this will identify which link set to get modify .i.e
$url= "link=$row['linkname']&linkurl=$row['link']";
<a href="samescript.name?$url"/>update</a>
so when you click on update link it will hit the same script and perform the update query
if(isset($_GET['link']) && isset($_GET['linkname']))
{
query to update
}
I hope this may help. Let me know if any issues you come across.
Use an image while querying news items.
in PHP Coding Help
Posted
This can be handled via css . you can ask take help in CSS section or some designer.